1963 TMTA High School Algebra I Contest Problem 16

Problem

The factors of $\frac{4}{9} + \frac{1}{3}x + \frac{3}{16}x^2$ are

$\text{(A)} \quad (1/9-x/16)(4-3x) \quad \text{(B)} \quad (2/9-x/4)(2-3x/4) \quad \text{(C)} \quad (4/9-3x/16)(1-x)$

$\text{(D)} \quad -(2/3-x/4)(2/3+3x/4) \quad \text{(E)} \quad 4/9-3x(1/16+1)$

Solution

We can immediately eliminate choice E, because this doesn't yield a quadratic. The other answer choices have constants that multiply to $\frac{4}{9},$ so all we care about is the middle $\frac{1}{3}x$ term.

Expand A to get $\frac{4}{9}-\frac{7}{12}x + \frac{3}{16}x^2.$

Expand B to get $\frac{4}{9}-\frac{2}{3}x + \frac{3}{16}x^2.$

Expand C to get $\frac{4}{9}-\frac{91}{144}x + \frac{3}{16}x^2.$

Expand D to get $\frac{4}{9}-\frac{1}{3}x + \frac{3}{16}x^2.$

We see that the answer is $\boxed{\text{(D)}}.$

1963 TMTA High School Mathematics Contests (Problems)
Preceded by Problem 15	TMTA High School Mathematics Contest Past Problems/Solutions	Followed by Problem 17

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1963 TMTA High School Algebra I Contest Problem 16

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